One person enters the stadium the first mimute, two people the second mimute, three people the third minute, and so on. In how many mimutes will there be 55 people in the stadium?
One person enters the stadium the first mimute, two people the second mimute, three people the third minute, and so on. In how many mimutes will there be 55 people in the stadium?
Hello,Originally Posted by maggie
you want to know the number of summands in a sum:
$\displaystyle 1+2+3+...+n=55\Longrightarrow \frac{1}{2}\cdot n\cdot (n+1)=55$
Expand the LHS of this equation and solve the quadratic equation for n:
$\displaystyle \frac{1}{2}n^2+\frac{1}{2}n-55=0$
You'll get x = -11 or x = 10. (-11) persons are a very ugly sight.
So the only possible solution for your problem is: 10 persons.
Greetings
EB